Let $\mathscr{O}(U)$ be a ring from a set $U$. Let $p \in U$, such that $\mathscr{O}_{p}$ is the stalk at $p$. If we consider $\mathscr{m}_{p}$ as the germs vanishing at $p$, then $\mathscr{m}_{p}$ is an ideal.
Show that $\mathscr{m}_{p}$ is the only maximal ideal of $\mathscr{O}_{p}$.
I was trying to show first that every element of $\mathscr{O}_{p} \backslash \mathscr{m}_{p}$ is invertible, since $\mathscr{O}_{p} \backslash \mathscr{m}_{p}$ is a field. But I don't get success in this.
Your definition is sketchy and incomplete. But the notation being traditional I am assuming $\mathcal{O}(U)$ means ring of continuous functions defined on an open set $U$ of a fixed space $X$ (could be differentiable, or holomorphic, or algebraic functions). Then show that any function $f$ not in $\mathcal{m}_p$ is invertible. As the function is non-zero at $p$ by continuity it will be non-zero in an open neighbourhood of $p$. And so $1/f$ will be well-defined and continuous(or hololorphic, algebraic) in that same neighbourhood. So we get an inverse in the ring, showing $\mathcal{O}_p$ is a local ring.